cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A238363 Coefficients for the commutator for the logarithm of the derivative operator [log(D),x^n D^n]=d[(xD)!/(xD-n)!]/d(xD) expanded in the operators :xD:^k.

Original entry on oeis.org

1, -1, 2, 2, -3, 3, -6, 8, -6, 4, 24, -30, 20, -10, 5, -120, 144, -90, 40, -15, 6, 720, -840, 504, -210, 70, -21, 7, -5040, 5760, -3360, 1344, -420, 112, -28, 8, 40320, -45360, 25920, -10080, 3024, -756, 168, -36, 9, -362880, 403200, -226800, 86400, -25200, 6048, -1260, 240, -45, 10
Offset: 1

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Author

Tom Copeland, Feb 25 2014

Keywords

Comments

Let D=d/dx and [A,B]=A·B-B·A. Then each row corresponds to the coefficients of the operators :xD:^k = x^k D^k in the expansion of the commutator [log(D),:xD:^n]=[-log(x),:xD:^n]=sum(k=0 to n-1, a(n,k) :xD:^k). The e.g.f. is derived from [log(D), exp(t:xD:)]=[-log(x), exp(t:xD:)]= log(1+t)exp(t:xD:), using the shift property exp(t:xD:)f(x)=f((1+t)x).
The reversed unsigned array is A111492.
See the mathoverflow link and link therein to an associated mathstackexchange question for other formulas for log(D). In addition, R_x = log(D) = -log(x) + c - sum[n=1 to infnty, (-1)^n 1/n :xD:^n/n!]=
-log(x) + Psi(1+xD) = -log(x) + c + Ein(:xD:), where c is the Euler-Mascheroni constant, Psi(x), the digamma function, and Ein(x), a breed of the exponential integrals (cf. Wikipedia). The :xD:^k ops. commute; therefore, the commutator reduces to the -log(x) term.
Also the n-th row corresponds to the expansion of d[(xD)!/(xD-n)!]/d(xD) = d[:xD:^n]/d(xD) in the operators :xD:^k, or, equivalently, the coefficients of x in d[z!/(z-n)!]/dz=d[St1(n,z)]]/dz evaluated umbrally with z=St2(.,x), i.e., z^n replaced by St2(n,x), where St1(n,x) and St2(n,x) are the signed and unsigned Stirling polynomials of the first (A008275) and second (A008277) kinds. The derivatives of the unsigned St1 are A028421. See examples. This formalism follows from the relations between the raising and lowering operators presented in the MathOverflow link and the Pincherle derivative. The results can be generalized through the operator relations in A094638, which are related to the celebrated Witt Lie algebra and pseudodifferential operators / symbols, to encompass other integral arrays.
A002741(n)*(-1)^(n+1) (row sums), A002104(n)*(-1)^(n+1) (alternating row sums). Column sequences: A133942(n-1), A001048(n-1), A238474, ... - Wolfdieter Lang, Mar 01 2014
Add an additional head row of zeros to the lower triangular array and denote it as T (with initial indexing in columns and rows being 0). Let dP = A132440, the infinitesimal generator for the Pascal matrix, and I, the identity matrix, then exp(T)=I+dP, i.e., T=log(I+dP). Also, (T_n)^n=0, where T_n denotes the n X n submatrix, i.e., T_n is nilpotent of order n. - Tom Copeland, Mar 01 2014
Any pair of lowering and raising ops. L p(n,x) = n·p(n-1,x) and R p(n,x) = p(n+1,x) satisfy [L,R]=1 which implies (RL)^n = St2(n,:RL:), and since (St2(·,u))!/(St2(·,u)-n)!= u^n, when evaluated umbrally, d[(RL)!/(RL-n)!]/d(RL) = d[:RL:^n]/d(RL) is well-defined and gives A238363 when the LHS is reduced to a sum of :RL:^k terms, exactly as for L=d/dx and R=x above. (Note that R_x above is a raising op. different from x, with associated L_x=-xD.) - Tom Copeland, Mar 02 2014
For relations to colored forests, disposition of flags on flagpoles, and the colorings of the vertices of the complete graphs K_n, encoded in their chromatic polynomials, see A130534. - Tom Copeland, Apr 05 2014
The unsigned triangle, omitting the main diagonal, gives A211603. See also A092271. Related to the infinitesimal generator of A008290. - Peter Bala, Feb 13 2017

Examples

			The first few row polynomials are
p(1,x)=  1
p(2,x)= -1 + 2x
p(3,x)=  2 - 3x + 3x^2
p(4,x)= -6 + 8x - 6x^2 + 4x^3
p(5,x)= 24 -30x +20x^2 -10x^3 + 5x^4
...........
For n=3: z!/(z-3)!=z^3-3z^2+2z=St1(3,z) with derivative 3z^2-6z+2, and
3·St2(2,x)-6·St2(1,x)+2=3(x^2+x)-6x+2=3x^2-3x+2=p(3,x). To see the relation to the operator formalism, note that (xD)^k=St2(k,:xD:) and (xD)!/(xD-k)!=[St2(·,:xD:)]!/[St2(·,:xD:)-k]!= :xD:^k.
The triangle a(n,k) begins:
n\k       0       1       2      3      4     5      6    7   8   9 ...
1:        1
2:       -1       2
3:        2      -3       3
4:       -6       8      -6      4
5:       24     -30      20    -10      5
6:     -120     144     -90     40    -15     6
7:      720    -840     504   -210     70   -21      7
8:    -5040    5760   -3360   1344   -420   112    -28    8
9:    40320  -45360   25920 -10080   3024  -756    168  -36   9
10: -362880  403200 -226800  86400 -25200  6048  -1260  240 -45  10
... formatted by _Wolfdieter Lang_, Mar 01 2014
-----------------------------------------------------------------------
		

Crossrefs

Programs

  • Mathematica
    a[n_, k_] := (-1)^(n-k-1)*n!/((n-k)*k!); Table[a[n, k], {n, 1, 10}, {k, 0, n-1}] // Flatten (* Jean-François Alcover, Jul 09 2015 *)

Formula

a(n,k) = (-1)^(n-k-1)*n!/((n-k)*k!) for k=0 to (n-1).
E.g.f.: log(1+t)*exp(x*t).
E.g.f.for unsigned array: -log(1-t)*exp(x*t).
The lowering op. for the row polynomials is L=d/dx, i.e., L p(n,x) = n*p(n-1,x).
An e.g.f. for an unsigned related version is -log(1+t)*exp(x*t)/t= exp(t*s(·,x)) with s(n,x)=(-1)^n * p(n+1,-x)/(n+1). Let L=d/dx and R= x-(1/((1-D)log(1-D))+1/D),then R s(n,x)= s(n+1,x) and L s(n,x)= n*s(n-1,x), defining a special Sheffer sequence of polynomials, an Appell sequence. So, R (-1)^(n-1) p(n,-x)/n = (-1)^n p(n+1,-x)/(n+1).
From Tom Copeland, Apr 17 2014: (Start)
Dividing each diagonal by its first element (-1)^(n-1)*(n-1)! yields the reverse of A104712.
Multiply each n-th diagonal of the Pascal lower triangular matrix by x^n and designate the result as A007318(x) = P(x). Then with dP = A132440, M = padded A238363 = A238385-I, I = identity matrix, and (B(.,x))^n = B(n,x) = the n-th Bell polynomial Bell(n,x) of A008277,
A) P(x)= exp(x*dP) = exp[x*(e^M-I)] = exp[M*B(.,x)] = (I+dP)^B(.,x), and
B) P(:xD:)=exp(dP:xD:)=exp[(e^M-I):xD:]=exp[M*B(.,:xD:)]=exp[M*xD]=
(1+dP)^(xD) with action P(:xD:)g(x) = exp(dP:xD:)g(x) = g[(I+dP)*x].
C) P(x)^m = P(m*x). P(2x) = A038207(x) = exp[M*B(.,2x)], face vectors of n-D hypercubes. (End)
From Tom Copeland, Apr 26 2014: (Start)
M = padded A238363 = A238385-I
A) = [St1]*[dP]*[St2] = [padded A008275]*A132440*A048993
B) = [St1]*[dP]*[St1]^(-1)
C) = [St2]^(-1)*[dP]*[St2]
D) = [St2]^(-1)*[dP]*[St1]^(-1),
where [St1]=padded A008275 just as [St2]=A048993=padded A008277.
E) P(x) = [St2]*exp(x*M)*[St1] = [St2]*(I + dP)^x*[St1].
F) exp(x*M) = [St1]*P(x)*[St2] = (I + dP)^x,
where (I + dP)^x = sum(k>=0, C(x,k)*dP^k).
Let the row vector Rv=(c0 c1 c2 c3 ...) and the column vector Cv(x)=(1 x x^2 x^3 ...)^Transpose. Form the power series V(x)= Rv * Cv(x) and W(y) := V(x.) evaluated umbrally with (x.)^n = x_n = (y)_n = y!/(y-n)!. Then
G) U(:xD:) = dV(:xD:)/d(xD) = dW(xD)/d(xD) evaluated with (xD)^n = Bell(n,:xD:),
H) U(x) = dV(x.)/dy := dW(y)/dy evaluated with y^n=y_n=Bell(n,x), and
I) U(x) = Rv * M * Cv(x). (Cf. A132440, A074909.) (End)
The Bernoulli polynomials Ber_n(x) are related to the polynomials q_n(x) = p(n+1,x) / (n+1) with the e.g.f. [log(1+t)/t] e^(xt) (cf. s_n (x) above) as Ber_n(x) = St2_n[q.(St1.(x))], umbrally, or [St2]*[q]*[St1], in matrix form. Since q_n(x) is an Appell sequence of polynomials, q_n(x) = [log(1+D_x)/D_x]x^n. - Tom Copeland, Nov 06 2016

Extensions

Pincherle formalism added by Tom Copeland, Feb 27 2014

A001721 Generalized Stirling numbers.

Original entry on oeis.org

1, 11, 107, 1066, 11274, 127860, 1557660, 20355120, 284574960, 4243508640, 67285058400, 1131047366400, 20099588140800, 376612896038400, 7422410595801600, 153516757766400000, 3325222830101760000, 75283691519393280000, 1778358268603445760000
Offset: 0

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The asymptotic expansion of the higher order exponential integral E(x,m=2,n=5) ~ exp(-x)/x^2*(1 - 11/x + 107/x^2 - 1066/x^3 + 11274/x^4 - 127860/x^5 + 1557660/x^6 - ... ) leads to the sequence given above. See A163931 and A028421 for more information. - Johannes W. Meijer, Oct 20 2009

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Related to n!*(the k-th successive summation of the harmonic numbers): k=0..A000254, k=1..A001705,k= 2..A001711, k=3..A001716, k=4..A001721, k=5..A051524, k=6..A051545, k=7..A051560, k=8..A051562, k=9..A051564.

Programs

  • Mathematica
    f[k_] := k + 4; t[n_] := Table[f[k], {k, 1, n}]; a[n_] := SymmetricPolynomial[n - 1, t[n]]; Table[a[n], {n, 1, 16}] (* Clark Kimberling, Dec 29 2011 *)

Formula

a(n) = Sum_{k=0..n} (-1)^(n+k)*binomial(k+1, 1)*5^k*Stirling1(n+1, k+1). - Borislav Crstici (bcrstici(AT)etv.utt.ro), Jan 26 2004
a(n) = n!*Sum_{k=0..n-1} (-1)^k*binomial(-5,k)/(n-k). - Milan Janjic, Dec 14 2008
a(n) = n!*[4]h(n), where [k]h(n) denotes the k-th successive summation of the harmonic numbers from 0 to n. With offset 1. - Gary Detlefs, Jan 04 2011
E.g.f.: (1 + 5*log(1/(1-x)))/(1 - x)^6. - Ilya Gutkovskiy, Jan 23 2017

Extensions

More terms from Borislav Crstici (bcrstici(AT)etv.utt.ro), Jan 26 2004

A051524 Second unsigned column of triangle A051338.

Original entry on oeis.org

0, 1, 13, 146, 1650, 19524, 245004, 3272688, 46536624, 703404576, 11277554400, 191338156800, 3427105248000, 64651956364800, 1281740285145600, 26648514872985600, 579892995734169600, 13183403757582643200
Offset: 0

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The asymptotic expansion of the higher order exponential integral E(x,m=2,n=6) ~ exp(-x)/x^2*(1 - 13/x + 146/x^2 - 1650/x^3 + 19524/x^4 - 245004/x^5 + 3272688/x^6 - ...) leads to the sequence given above. See A163931 and A028421 for more information. - Johannes W. Meijer, Oct 20 2009

References

  • Mitrinovic, D. S. and Mitrinovic, R. S.: see reference given for triangle A051338.

Crossrefs

Cf. A001725 (first unsigned column).
Related to n!*the k-th successive summation of the harmonic numbers: k=0..A000254, k=1..A001705, k= 2..A001711, k=3..A001716, k=4..A001721, k=5..A051524, k=6..A051545, k=7..A051560, k=8..A051562, k=9..A051564. - Gary Detlefs, Jan 04 2011

Programs

  • Mathematica
    f[k_] := k + 5; t[n_] := Table[f[k], {k, 1, n}]
    a[n_] := SymmetricPolynomial[n - 1, t[n]]
    Table[a[n], {n, 1, 16}]
    (* Clark Kimberling, Dec 29 2011 *)

Formula

a(n) = A051338(n, 1)*(-1)^(n-1);
E.g.f.: -log(1-x)/(1-x)^6.
For n>=1, a(n) = n!*Sum_{k=0..n-1} (-1)^k*binomial(-6,k)/(n-k). - Milan Janjic, Dec 14 2008
a(n) = n!*[5]h(n), where [k]h(n) denotes the k-th successive summation of h(n) from 0 to n. - Gary Detlefs, Jan 04 2011
Conjecture: a(n) +(-2*n-9)*a(n-1) +(n+4)^2*a(n-2)=0. - R. J. Mathar, Aug 04 2013

A051545 Second unsigned column of triangle A051339.

Original entry on oeis.org

0, 1, 15, 191, 2414, 31594, 434568, 6314664, 97053936, 1576890000, 27046454400, 488849155200, 9293295110400, 185464792800000, 3878247384345600, 84822225638169600, 1937048605944883200, 46113230058645657600
Offset: 0

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Comments

The asymptotic expansion of the higher order exponential integral E(x,m=2,n=7) ~ exp(-x)/x^2*(1 - 15/x + 191/x^2 - 2414/x^3 + 31594/x^4 - 434568/x^5 + 6314664/x^6 - ...) leads to the sequence given above. See A163931 and A028421 for more information. - Johannes W. Meijer, Oct 20 2009

References

  • Mitrinovic, D. S. and Mitrinovic, R. S. see reference given for triangle A051339.

Crossrefs

Cf. A001730 (first unsigned column).
Related to n!*the k-th successive summation of the harmonic numbers: k=0..A000254, k=1..A001705, k= 2..A001711, k=3..A001716, k=4..A001721, k=5..A051524, k=6..(this sequence), k=7..A051560, k=8..A051562, k=9..A051564. - Gary Detlefs, Jan 04 2011

Programs

  • Mathematica
    f[k_] := k + 6; t[n_] := Table[f[k], {k, 1, n}]
    a[n_] := SymmetricPolynomial[n - 1, t[n]]
    Table[a[n], {n, 1, 16}]
    (* Clark Kimberling, Dec 29 2011 *)

Formula

a(n) = A051339(n, 2)*(-1)^(n-1).
E.g.f.: -log(1-x)/(1-x)^7.
a(n) = n!*Sum_{k=0,..,n-1}((-1)^k*binomial(-7,k)/(n-k)), for n>=1. - Milan Janjic, Dec 14 2008
a(n) = n!*[6]h(n), where [k]h(n) denotes the k-th successive summation of The harmonic numbers from 0 to n. - Gary Detlefs, Jan 04 2011

A051560 Second unsigned column of triangle A051379.

Original entry on oeis.org

0, 1, 17, 242, 3382, 48504, 725592, 11393808, 188204400, 3270729600, 59753750400, 1146140409600, 23046980025600, 485075533132800, 10669304848204800, 244861798361241600, 5854837379724748800
Offset: 0

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The asymptotic expansion of the higher order exponential integral E(x,m=2,n=8) ~ exp(-x)/x^2*(1 - 17/x + 242/x^2 - 3382/x^3 + 48504/x^4 - 725592/x^5 + 11393808/x^6 - ...) leads to the sequence given above. See A163931 and A028421 for more information. - Johannes W. Meijer, Oct 20 2009

References

  • Mitrinovic, D. S. and Mitrinovic, R. S. see reference given for triangle A051379.

Crossrefs

Cf. A049388 (first unsigned column).
Related to n!*the k-th successive summation of the harmonic numbers: k=0..A000254, k=1..A001705, k= 2..A001711, k=3..A001716, k=4..A001721, k=5..A051524, k=6..A051545, k=7..A051560, k=8..A051562, k=9..A051564. - Gary Detlefs Jan 04 2011

Programs

  • Mathematica
    f[k_] := k + 7; t[n_] := Table[f[k], {k, 1, n}]
    a[n_] := SymmetricPolynomial[n - 1, t[n]]
    Table[a[n], {n, 1, 16}]
    (* Clark Kimberling, Dec 29 2011 *)

Formula

a(n) = A051379(n, 2)*(-1)^(n-1).
E.g.f.: -log(1-x)/(1-x)^8.
a(n) = n!*Sum_{k=0..n-1} ((-1)^k*binomial(-8,k)/(n-k)), for n>=1. - Milan Janjic, Dec 14 2008
a(n) = n!*[7]h(n), where [k]h(n) denotes the k-th successive summation of the harmonic numbers from 0 to n. - Gary Detlefs, Jan 04 2011
Conjecture: a(n) +(-2*n-13)*a(n-1) +(n+6)^2*a(n-2)=0. - R. J. Mathar, Aug 04 2013

A080663 a(n) = 3*n^2 - 1.

Original entry on oeis.org

2, 11, 26, 47, 74, 107, 146, 191, 242, 299, 362, 431, 506, 587, 674, 767, 866, 971, 1082, 1199, 1322, 1451, 1586, 1727, 1874, 2027, 2186, 2351, 2522, 2699, 2882, 3071, 3266, 3467, 3674, 3887, 4106, 4331, 4562, 4799, 5042, 5291, 5546, 5807, 6074, 6347, 6626
Offset: 1

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Author

Cino Hilliard, Mar 01 2003

Keywords

Comments

These numbers cannot be perfect squares. See the Hilliard link for a proof.
2nd elementary symmetric polynomial of n, n + 1 and n + 2: n(n+1) + n(n+2) + (n+1)(n+2). - Zak Seidov, Mar 23 2005
This sequence equals for n >= 2 the third right hand column of triangle A165674. Its recurrence relation leads to Pascal's triangle A007318. Crowley's formula for A080663(n-1) leads to Wiggen's triangle A028421 and the o.g.f. of this sequence, without the first term, leads to Wood's polynomials A126671. See also A165676, A165677, A165678 and A165679. - Johannes W. Meijer, Oct 16 2009
The Diophantine equation x(x+1) + (x+2)(x+3) = (x+y)^2 + (x-y)^2 has solutions x = a(n), y = 3n. - Bruno Berselli, Mar 29 2013
A simpler proof that these numbers can't be perfect squares can easily be constructed using congruences: If the equation x^2 = 3y^2 - 1 has a solution in positive integers, then x^2 = 2 mod 3. Obviously we can't have x = 0 mod 3, and x = 1 mod 3 doesn't work either because then x^2 = 1 mod 3 also. That leaves x = 2 mod 3, but then x^2 = 1 mod 3. - Alonso del Arte, Oct 19 2013
2*a(n+1) is surface area of a rectangular prism with consecutive integer sides: n, n+1, and n+2, (n>0). - Wesley Ivan Hurt, Sep 06 2014
Numbers m such that 3*m+3 is a square. So, these are the numbers m such that the system of equations x=sqrt(m-2yz), y=sqrt(m+1-2xz), z=sqrt(m+2-2xy) admits 3 real positive solutions whose sum is an integer. See the Rechtman link. - Michel Marcus, Jun 06 2020

References

  • Ethan D. Bolker, Elementary Number Theory: An Algebraic Approach. Mineola, New York: Dover Publications (1969, reprinted 2007): p. 7, Problem 6.6.
  • E. Grosswald, Topics from the Theory of Numbers, 1966 p 64 problem 11

Crossrefs

Programs

Formula

a(n) = -Re((1 + n*i)^3) where i=sqrt(-1). - Gary W. Adamson, Aug 14 2006
a(n) = 3*n^2 - 1. - Stephen Crowley, Jul 06 2009
a(n) = a(n-1) - 3*a(n-2) + 3*a(n-3). - Johannes W. Meijer, Oct 16 2009
G.f.: x*(2 + 5*x - x^2)/(1-x)^3. - Joerg Arndt, Sep 06 2014
a(n) = a(n-1) + 6*n - 3 for n > 1. - Vincenzo Librandi, Aug 08 2010
E.g.f.: 1 + exp(x)*(3*x^2 + 3*x - 1). - Stefano Spezia, Feb 01 2020
From Amiram Eldar, Feb 04 2021: (Start)
Sum_{n>=1} 1/a(n) = (1 - (Pi/sqrt(3))*cot(Pi/sqrt(3)))/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = ((Pi/sqrt(3))*csc(Pi/sqrt(3)) - 1)/2.
Product_{n>=1} (1 + 1/a(n)) = (Pi/sqrt(3))*csc(Pi/sqrt(3)).
Product_{n>=1} (1 - 1/a(n)) = csc(Pi/sqrt(3))*sin(sqrt(2/3)*Pi)/sqrt(2). (End)
a(n) = A005449(n) + A115067(n). - Leo Tavares, May 25 2022
a(n) = (n-1)*n + (n-1)*(n+1) + n*(n+1), for n >= 1. See the Zak Seidov comment above. - Wolfdieter Lang, Aug 15 2024

A051562 Second unsigned column of triangle A051380.

Original entry on oeis.org

0, 1, 19, 299, 4578, 71394, 1153956, 19471500, 343976400, 6366517200, 123418922400, 2503748556000, 53091962697600, 1175271048201600, 27123099523027200, 651708291206649600, 16282170039031142400
Offset: 0

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Comments

The asymptotic expansion of the higher order exponential integral E(x,m=2,n=9) ~ exp(-x)/x^2*(1 - 19/x + 299/x^2 - 4578/x^3 + 71394/x^4 - 1153956/x^5 + 19471500/x^6 - ...) leads to the sequence given above. See A163931 and A028421 for more information. - Johannes W. Meijer, Oct 20 2009

References

  • Mitrinovic, D. S. and Mitrinovic, R. S. see reference given for triangle A051380.

Crossrefs

Cf. A049389 (first unsigned column).
Related to n!*the k-th successive summation of the harmonic numbers: k=0..A000254, k=1..A001705, k= 2..A001711, k=3..A001716, k=4..A001721, k=5..A051524, k=6..A051545, k=7..A051560, k=8..A051562, k=9..A051564. - Gary Detlefs Jan 04 2011

Programs

  • Mathematica
    f[k_] := k + 8; t[n_] := Table[f[k], {k, 1, n}]
    a[n_] := SymmetricPolynomial[n - 1, t[n]]
    Table[a[n], {n, 1, 16}]
    (* Clark Kimberling, Dec 29 2011 *)

Formula

a(n) = A051380(n, 2)*(-1)^(n-1).
E.g.f.: -log(1-x)/(1-x)^9.
a(n) = n!*Sum_{k=0..n-1} ((-1)^k*binomial(-9,k)/(n-k)), for n>=1. - Milan Janjic, Dec 14 2008
a(n) = n!*[8]h(n), where [k]h(n) denotes the k-th successive summation of the harmonic numbers from 0 to n. - Gary Detlefs, Jan 04 2011

A051564 Second unsigned column of triangle A051523.

Original entry on oeis.org

0, 1, 21, 362, 6026, 101524, 1763100, 31813200, 598482000, 11752855200, 240947474400, 5154170774400, 114942011990400, 2669517204076800, 64496340380102400, 1619153396908185600, 42188624389562112000
Offset: 0

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Author

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Comments

The asymptotic expansion of the higher order exponential integral E(x,m=2,n=10) ~ exp(-x)/x^2*(1 - 21/x + 362/x^2 - 6026/x^3 + 101524/x^4 - 1763100/x^5 + 31813200/x^6 - ...) leads to the sequence given above. See A163931 and A028421 for more information. - Johannes W. Meijer, Oct 20 2009

References

  • Mitrinovic, D. S. and Mitrinovic, R. S. see reference given for triangle A051523.

Crossrefs

Cf. A049398 (first unsigned column).
Related to n!*the k-th successive summation of the harmonic numbers: k=0..A000254, k=1..A001705, k=2..A001711, k=3..A001716, k=4..A001721, k=5..A051524, k=6..A051545, k=7..A051560, k=8..A051562, k=9..A051564. - Gary Detlefs Jan 04 2011

Programs

  • Mathematica
    f[n_] := n!*Sum[(-1)^k*Binomial[-10, k]/(n - k), {k, 0, n - 1}]; Array[f, 17, 0]
    Range[0, 16]! CoefficientList[ Series[-Log[(1 - x)]/(1 - x)^10, {x, 0, 16}], x]
    (* Or, using elementary symmetric functions: *)
    f[k_] := k + 9; t[n_] := Table[f[k], {k, 1, n}]
    a[n_] := SymmetricPolynomial[n - 1, t[n]]
    Table[a[n], {n, 1, 16}]
    (* Clark Kimberling, Dec 29 2011 *)

Formula

a(n) = A051523(n, 2)*(-1)^(n-1).
E.g.f.: -log(1-x)/(1-x)^10.
a(n) = n!*Sum_{k=0..n-1}((-1)^k*binomial(-10,k)/(n-k)), for n>=1. - Milan Janjic, Dec 14 2008
a(n) = n!*[9]h(n), where [k]h(n) denotes the k-th successive summation of the harmonic numbers from 0 to n. - Gary Detlefs, Jan 04 2011

A061356 Triangle read by rows: T(n, k) is the number of labeled trees on n nodes with maximal node degree k (0 < k < n).

Original entry on oeis.org

1, 2, 1, 9, 6, 1, 64, 48, 12, 1, 625, 500, 150, 20, 1, 7776, 6480, 2160, 360, 30, 1, 117649, 100842, 36015, 6860, 735, 42, 1, 2097152, 1835008, 688128, 143360, 17920, 1344, 56, 1, 43046721, 38263752, 14880348, 3306744, 459270, 40824, 2268, 72, 1
Offset: 2

Views

Author

Olivier Gérard, Jun 07 2001

Keywords

Comments

Essentially the coefficients of the Abel polynomials (A137452). - Peter Luschny, Jun 12 2022
This is a formula from Comtet, Theorem F, vol. I, p. 81 (French edition) used in proving Theorem D.
If we let N = n+1, binomial(N-2, k-1)*(N-1)^(N-k-1) = binomial(n-1, k-1)*n^(n-k), so this sequence with offset 1,1 also gives the number of rooted forests of k trees over [n]. - Washington Bomfim, Jan 09 2008
Let S(n,k) be the signed triangle, S(n,k) = (-1)^(n-k)T(n,k), which starts 1, -2, 1, 9, -6, 1, ..., then the inverse of S is the triangle of idempotent numbers A059298. - Peter Luschny, Mar 13 2009
With offset 1 also number of labeled multigraphs of k components, n nodes, and no cycles except one loop in each component. See link below to have a picture showing the bijection between rooted forests and multigraphs of this kind. (Note that there are no labels in the picture, but the bijection remains true if we label the nodes.) - Washington Bomfim, Sep 04 2010
With offset 1, T(n,k) is the number of forests of rooted trees on n nodes with exactly k (rooted) trees. - Geoffrey Critzer, Feb 10 2012
Also the Bell transform of the sequence (n+1)^n (A000169(n+1)) without column 0. For the definition of the Bell transform see A264428. - Peter Luschny, Jan 21 2016
Abel polynomials A(n,x) = x*(x+n)^(n-1) satisfy d/dx A(n,x) = n*A(n-1,x+1). - Michael Somos, May 10 2024
Also, T(n,k) is the number of parking functions with k ties. - Kyle Celano, Aug 18 2025

Examples

			Triangle begins
    1;
    2,     1;
    9,     6,     1;
   64,    48,    12,    1;
  625,   500,   150,   20,    1;
 7776,  6480,  2160,  360,   30,    1;
 ...
From _Peter Bala_, Sep 21 2012: (Start)
O.g.f.'s for the diagonals begin:
1/(1-x) = 1 + x + x^2 + x^3 + ...
2*x/(1-x)^3 = 2 + 6*x + 12*x^3 + ... A002378(n+1)
(9+3*x)/(1-x)^5 = 9 + 48*x + 150*x^2 + ... 3*A004320(n+1)
The numerator polynomials are the row polynomials of A155163.
(End)
		

References

  • L. Comtet, Analyse Combinatoire, P.U.F., Paris 1970. Volume 1, p 81.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974.

Crossrefs

Variant of A137452.
First diagonal is A002378.
Row sums give A000272.
Cf. A028421, A059297, A139526 (row reverse), A155163, A202017.

Programs

  • Maple
    # The function BellMatrix is defined in A264428.
    # Adds (1,0,0,0,...) as column 0 to the triangle.
    BellMatrix(n -> (n+1)^n, 12); # Peter Luschny, Jan 21 2016
  • Mathematica
    nn = 7; t = Sum[n^(n - 1)  x^n/n!, {n, 1, nn}]; f[list_] := Select[list, # > 0 &]; Map[f, Drop[Range[0, nn]! CoefficientList[Series[Exp[y t], {x, 0, nn}], {x, y}], 1]] // Flatten  (* Geoffrey Critzer, Feb 10 2012 *)
    T[n_, m_] := T[n, m] = Binomial[n, m]*Sum[m^k*T[n-m, k], {k, 1, n-m}]; T[n_, n_] = 1; Table[T[n, m], {n, 1, 9}, {m, 1, n}] // Flatten (* Jean-François Alcover, Mar 31 2015, after Vladimir Kruchinin *)
    Table[Binomial[n - 2, k - 1]*(n - 1)^(n - k - 1), {n, 2, 12}, {k, 1, n - 1}] // Flatten (* G. C. Greubel, Nov 12 2017 *)
    BellMatrix[f_Function, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len-1}, {k, 0, len-1}]];
    rows = 10;
    M = BellMatrix[(# + 1)^#&, rows];
    Table[M[[n, k]], {n, 2, rows}, {k, 2, n}] // Flatten (* Jean-François Alcover, Jun 23 2018, after Peter Luschny *)
  • Maxima
    create_list(binomial(n,k)*(n+1)^(n-k),n,0,20,k,0,n); /* Emanuele Munarini, Apr 01 2014 */
    
  • PARI
    for(n=2,11, for(k=1,n-1, print1(binomial(n-2, k-1)*(n-1)^(n-k-1), ", "))) \\ G. C. Greubel, Nov 12 2017
  • Sage
    # uses[bell_matrix from A264428]
    # Adds (1,0,0,0,...) as column 0 to the triangle.
    bell_matrix(lambda n: (n+1)^n, 12) # Peter Luschny, Jan 21 2016
    

Formula

T(n, k) = binomial(n-2, k-1)*(n-1)^(n-k-1).
E.g.f.: (-LambertW(-y)/y)^(x+1)/(1+LambertW(-y)). - Vladeta Jovovic
From Peter Bala, Sep 21 2012: (Start)
Let T(x) = Sum_{n >= 0} n^(n-1)*x^n/n! denote the tree function of A000169. E.g.f.: F(x,t) := exp(t*T(x)) - 1 = -1 + {T(x)/x}^t = t*x + t*(2 + t)*x^2/2! + t*(9 + 6*t + t^2)*x^3/3! + ....
The compositional inverse with respect to x of (1/t)*F(x,t) is the e.g.f. for a signed version of the row reverse of A028421.
The row generating polynomials are the Abel polynomials A(n,x) = x*(x+n)^(n-1) for n >= 1.
Define B(n,x) = x^n/(1+n*x)^(n+1) = (-1)^n*A(-n,-1/x) for n >= 1. The k-th column entries are the coefficients in the formal series expansion of x^k in terms of B(n,x). For example, Col. 1: x = B(1,x) + 2*B(2,x) + 9*B(3,x) + 64*B(4,x) + ..., Col. 2: x^2 = B(2,x) + 6*B(3,x) + 48*B(4,x) + 500*B(5,x) + ... Compare with A059297.
n-th row sum = A000272(n+1).
Row reverse triangle is A139526.
The o.g.f.'s for the diagonals of the triangle are the rational functions R(n,x)/(1-x)^(2*n+1), where R(n,x) are the row polynomials of A155163. See below for examples.
(End)
T(n,m) = C(n,m)*Sum_{k=1..n-m} m^k*T(n-m,k), T(n,n) = 1. - Vladimir Kruchinin, Mar 31 2015

A165674 Triangle generated by the asymptotic expansions of the E(x,m=2,n).

Original entry on oeis.org

1, 3, 1, 11, 5, 1, 50, 26, 7, 1, 274, 154, 47, 9, 1, 1764, 1044, 342, 74, 11, 1, 13068, 8028, 2754, 638, 107, 13, 1, 109584, 69264, 24552, 5944, 1066, 146, 15, 1, 1026576, 663696, 241128, 60216, 11274, 1650, 191, 17, 1
Offset: 1

Views

Author

Johannes W. Meijer, Oct 05 2009

Keywords

Comments

The higher order exponential integrals E(x,m,n) are defined in A163931. The asymptotic expansion of the E(x,m=2,n) ~ (exp(-x)/x^2)*(1 - (1+2*n)/x + (2+6*n+3*n^2)/x^2 - (6+22*n+18*n^2+ 4*n^3)/x^3 + ... ) is discussed in A028421. The formula for the asymptotic expansion leads for n = 1, 2, 3, .., to the left hand columns of the triangle given above.
The recurrence relations of the right hand columns of this triangle lead to Pascal's triangle A007318, their a(n) formulas lead to Wiggen's triangle A028421 and their o.g.f.s lead to Wood's polynomials A126671; cf. A080663, A165676, A165677, A165678 and A165679.
The row sums of this triangle lead to A093344. Surprisingly the e.g.f. of the row sums Egf(x) = (exp(1)*Ei(1,1-x) - exp(1)*Ei(1,1))/(1-x) leads to the exponential integrals in view of the fact that E(x,m=1,n=1) = Ei(n=1,x). We point out that exp(1)*Ei(1,1) = A073003.
The Maple programs generate the coefficients of the triangle given above. The first one makes use of a relation between the triangle coefficients, see the formulas, and the second one makes use of the asymptotic expansions of the E(x,m=2,n).
Amarnath Murthy discovered triangle A093905 which is the reversal of our triangle.
A165675 is an extended version of this triangle. Its reversal is A105954.
Triangle A094587 is generated by the asymptotic expansions of E(x,m=1,n).

Crossrefs

A093905 is the reversal of this triangle.
A000254, A001705, A001711, A001716, A001721, A051524, A051545, A051560, A051562, A051564 are the first ten left hand columns.
A080663, n>=2, is the third right hand column.
A165676, A165677, A165678 and A165679 are the next right hand columns, A093344 gives the row sums.
A073003 is Gompertz's constant.
A094587 is generated by the asymptotic expansions of E(x, m=1, n).
Cf. A165675, A105954 (Quet) and A067176 (Bottomley).
Cf. A007318 (Pascal), A028421 (Wiggen), A126671 (Wood).

Programs

  • Maple
    nmax:=9; for n from 1 to nmax do a(n, n) := 1 od: for n from 2 to nmax do a(n, 1) := n*a(n-1, 1) + (n-1)! od: for n from 3 to nmax do for m from 2 to n-1 do a(n, m) := (n-m+1)*a(n-1, m) + a(n-1, m-1) od: od: seq(seq(a(n, m), m = 1..n), n = 1..nmax);
    # End program 1
    nmax := nmax+1: m:=2; with(combinat): EA := proc(x, m, n) local E, i; E:=0: for i from m-1 to nmax+2 do E := E + sum((-1)^(m+k1+1) * binomial(k1, m-1) * n^(k1-m+1) * stirling1(i, k1), k1=m-1..i) / x^(i-m+1) od: E:= exp(-x)/x^(m) * E: return(E); end: for n1 from 1 to nmax do f(n1-1) := simplify(exp(x) * x^(nmax+3) * EA(x, m, n1)); for m1 from 0 to nmax+2 do b(n1-1, m1) := coeff(f(n1-1), x, nmax+2-m1) od: od: for n1 from 0 to nmax-1 do for m1 from 0 to n1-m+1 do a(n1-m+2, m1+1) := abs(b(m1, n1-m1)) od: od: seq(seq(a(n, m), m = 1..n),n = 1..nmax-1);
    # End program 2
    # Maple programs revised by Johannes W. Meijer, Sep 22 2012

Formula

a(n,m) = (n-m+1)*a(n-1,m) + a(n-1,m-1), for 2 <= m <= n-1, with a(n,n) = 1 and a(n,1) = n*a(n-1,1) + (n-1)!.
a(n,m) = product(i, i= m..n)*sum(1/i, i = m..n).
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